Discrete version of Dudley's inequalilty: Assuming set is finite wlog?

Question

I'm self-studying Vershynin's High-dimensional probability book. I have a question about the proof of Theorem 8.1.4:

Let $(X_t)_{t\in T}$ be a mean-zero random process on a metric space $(T,d)$, with sub-gaussian increments. Then: $$ \mathbb{E}\sup_{t\in T}X_t\leq C K \sum_{k\in \mathbb{Z}}2^{-k}\sqrt{\log\mathcal{N}(T,d,2^{-k})} $$ Here, $C$ is a constant and $\mathcal{N}(T,d,2^{-k})$ the covering number of the set $T$ with balls of radius $2^{-k}$.

Question: In the proof, it is stated that without loss of generality, we may assume that $T$ is finite. Why can we do that?

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By sub-gaussian increments, the following is meant:

Consider a random process $(x_t)_{t\in T}$ on a metric space $(T,d)$. We say that the process has sub-gaussian increments, if there exists a constant $K>0$, such that for all $t,x\in T$, $$ ||X_t-X_s||_{\psi_2}\leq Kd(t,s) $$ Here, $||\cdot||_{\psi_2}$ is the sub-gaussian norm.

Answer 1

I think I found an explanation: To avoid issues with measurability, Vershynin defines: $$ \mathbb{E}[\sup_{t\in T}X_t]:=\sup_{T'\in \mathcal{T}}\mathbb{E}[\sup_{t\in T'}X_t] $$ Where $\mathcal{T}$ is the set of finite subsets of $T$ (see footnote 3 on p.151). So, the supremum is only taken over finite sets $T'$. The proof in Theorem 8.1.4 for finite sets gives us: $$ \sup_{T'\in \mathcal{T}}\mathbb{E}[\sup_{t\in T'}X_t] \leq \sup_{T'\in \mathcal{T}}CK\sum_{k\in\mathbb{Z}}2^{-k}\sqrt{\log\mathcal{N}(T',d,2^{-k})} $$ And since $T'\subset T$, we have $\mathcal{N}(T',d,2^{-k})\leq \mathcal{N}(T,d,2^{-k})$, and so: $$ \leq CK\sum_{k\in\mathbb{Z}}2^{-k}\sqrt{\log\mathcal{N}(T,d,2^{-k})} $$

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I am still curious whether this result holds more generally, i.e. without defining $\mathbb{E}[\sup_{t\in T}X_t]$ as Vershynin does. On the other hand, $\sup_{t\in T}X_t$ may be non-measurable, which would mean that the inequality does not make sense in the first place.

Answer 2

It's not the exact answer to your question but I had same problem in Generic chaining problem and here is my solution to that problem

<(Generic chaining bound) Let $(X_t)_{t \in T}$ be a mean-zero random process on a infinite ($T$ is infinite set) metric space $(T, d)$ with sub-gaussian increments i.e. it's satisfy \begin{equation}\label{eq-sub-gaussian-incriment-metric} \lVert X_t - X_s \rVert_{\psi_2} \leq K d(t,s) \text{ for all } t,s \in T. \end{equation} Then \begin{equation} \label{eq-generic-chaining-bound} \sup_{T_0 \subset T, |T_0|<|N|} E \max_{t \in T_0} X_t \leq CK \gamma_2(T,d) \end{equation}

Which means that we want to prove $$ E \max_{t \in W} X_t \leq CK \gamma_2(T,d) $$ for every finite $W \subset T$.

remark

Let's consider some examples of spaces and their $\gamma_2$ functions. We only focus on standard euclidean norm, so in this remark we don't write, which norm we use.

$A = \{-1 , 0, 1\}, B = \{-1 , 1\} \subset R$ $$ \gamma_2(A) = 1 \leq 2 = \gamma_2(B) $$

$A = [-1;1], B = \{-1 , 1\} \subset R$ $$ 1 \leq \gamma_2(A) \leq 2 = \gamma_2(B) $$

$A = [-1;1], B = [-1;-0,5] \cup [0,5;1] \subset R$ $$ \gamma_2(A) \leq \gamma_2(B) $$

$A = [-1;1], B = [-2;2] \subset R$ $$ \gamma_2(A) \leq \gamma_2(B) $$

Remark show us that in theorem we can't simply say $$ E \max_{t \in W} X_t \leq CK \gamma_2(W,d) \leq CK \gamma_2(T,d) $$ so we need another tool.\

(Generic chaining bound)\ If $K=0$ then left hand equal to $0$ so inequality is trivial. If $K \neq 0$ and $\gamma_2(T,d) = +\infty$ then inequality is trivial in this case too. So we can assume by linearity of norms and expectation that $K=1$ and $\gamma_2(T,d) < +\infty$. Let's fix some finite $W \subset T$, there is $n \in N$ such that $|W| < 2^{2^n}$. Let's fix some $e >0$ too, there is $(V_n)_{n \in N}$ admissible sequences such that $|V_k| = 2^{2^k}$ $k = 1,2,...$ and $$ \gamma_2(T,d) + e = \inf_{(T_k)} \sup_{t \in T} \sum \limits_{k=0}^{\infty} 2^{k/2}d(t,T_k) +e > \sup_{t \in T} \sum \limits_{k=0}^{\infty} 2^{k/2}d(t,V_k) . $$ Denote by $W_k = V_k$ $k = 0,1,...,n$ and $W_{n+i} = W_n \cup W$ $i=1,2,...$ and use same theorem for $W_{n+1}$ (we can do it because $W_{n+1}$ is finite) \begin{multline*} E \max_{t \in W} \leq E \max_{t \in W_{n+1}} X_t \leq \\ C\gamma_2(W_{n+1},d) = C\inf_{(T_k)} \sup_{t \in W_{n+1}} \sum \limits_{k=0}^{\infty} 2^{k/2}d(t,T_k) \leq\\ C\sup_{t \in W_{n+1}} \sum \limits_{k=0}^{\infty} 2^{k/2}d(t,W_k) = C\sup_{t \in W_{n+1}} \sum \limits_{k=0}^{n} 2^{k/2}d(t,W_k) \leq \\ C\sup_{t \in T} \sum \limits_{k=0}^{n} 2^{k/2}d(t,W_k) = C\sup_{t \in T} \sum \limits_{k=0}^{n} 2^{k/2}d(t,V_k) \leq \\ C\sup_{t \in T} \sum \limits_{k=0}^{\infty} 2^{k/2}d(t,V_k)< C (\gamma_2(T,d) + e) $$ \end{multline*} Finally note that $\epsilon$ can be arbitrary small, which is complete the proof.